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Let $G$ be a finite group on $84$ elements. The size of a largest possible proper subgroup of $G$ is _____

42.....

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Order of a Subgroup always divides the order of Group.
Proper Subgroup of Group having order $84$ would have one of the order (proper factors of $84)$ $2,3, 4,6,7,12,14, 21,28, 42$.

So the largest order would be $42$.
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Will the number of subgroups possible be $\Phi (84)$ = 24 ?

$\Phi(84) = 24$, number of generator,it is valid on cyclic group.

So here trivial subgroups can be identity elements and the group itself. Right?
Yes

Lagrange's theorem states that order of every subgroup of G, it must be the divisor of G.

So the largest subgroup will be 84 which is trivial, but in the question it is asking for the proper subgroup hence it will be 42.

Reference: https://en.wikipedia.org/wiki/Subgroup

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got same
but for prime divisor we  are sure  that it would be proper subgroup for other divisor it may or not be proper subgroup converse of lagrange's theorem is not true
Order of Group must be divisible by order of subgroup = 42.
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42 is correct
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